**Use the binomial theorem to find the indicated coefficient or term :**

**Here we are going to see some examples to understand how to find the indicated coefficient or term using binomial theorem.**

**To find the particular term of the expansion, we need to use the formula given below.**

**General term :**

T_{(r+1)} = ^{n}c_{r} x^{(n-r)} a^{r}

**Example 1 :**

Find the coefficient of x^{5} in the expansion of (x + 1/x^{3})^{7}

**Solution :**

**T _{(r+1)} = ^{n}c_{r} x^{(n-r)} a^{r}**

Comparing the given expression with the form (x + a)^{n}, we get x = x, a = 1/x^{3} and n = 7

T_{(r+1)} = ^{17}c_{r} x^{(17-r)} (1/x^{3})^{r}

= ^{17}c_{r} x^{(17-r)} (x^{-3})^{r}

= ^{17}c_{r} x^{(17-r)} (x^{-3r})

= ^{17}c_{r} x^{(17-r-3r)}

= ^{17}c_{r} x^{(17-4r)}

Let T_{r + 1} be the term containing x^{5}

17 − 4r = 5 ⇒ r = 3

T_{r + 1} = T_{3 + 1}

= ^{17}C_{3} x^{17 − 4(3)}

= 680x^{5}

Hence the coefficient of x^{5 }= 680

**Example 2 :**

Find the coefficient of x^{5} in the expansion of (x - 1/x)^{11}

**Solution :**

**T _{(r+1)} = ^{n}c_{r} x^{(n-r)} a^{r}**

Comparing the given expression with the form (x + a)^{n}, we get x = x, a = -1/x and n = 11

T_{(r+1)} = ^{11}c_{r} x^{(11-r)} (-1/x)^{r}

= ^{11}c_{r} x^{(11-r)} (-x^{-1})^{r}

= ^{11}c_{r} x^{(11-r)} (-x^{-r})

= -^{11}c_{r} x^{(11-r-r)}

= - ^{11}c_{r} x^{(11-2r)}

Let T_{r + 1} be the term containing x^{5}

11 − 2r = 5 ⇒ r = 3

T_{r + 1} = T_{3 + 1}

= -^{11}C_{3} x^{11 − 2(3)}

^{11}C_{3 } = (11⋅10⋅9)/(3⋅2⋅1)

= -165x^{5}

Hence the coefficient of x^{5 }= -165

**Example 3 :**

Find the constant term in the expansion (√x - 2/x^{2})^{10}

**Solution :**

**T _{(r+1)} = ^{n}c_{r} x^{(n-r)} a^{r}**

Comparing the given expression with the form (x + a)^{n}, we get x = √x, a = -2/x^{2} and n = 10

T_{(r+1)} = ^{10}c_{r} √x^{(10-r)} (-2/x^{2})^{r}

= ^{10}c_{r} x^{1/2}^{(10-r)} (-2x^{-2})^{r}

= ^{10}c_{r} x^{(10-r)/2} (-2x^{-2})^{r}

= (-2)^{r} ^{10}c_{r} x^{(10-r)/2} x^{-2r}

= (-2)^{r} ^{10}c_{r} x^{(10-r-4r)/2}

= (-2)^{r} ^{10}c_{r} x^{(10-5r)/2}

Let T_{r + 1} be the constant term

(10 - 5r)/2 = 0 ⇒ r = 2

T_{r + 1} = T_{2 + 1}

= (-2)^{2} ^{10}c_{2} x^{(10-5(2))/2}

^{10}C_{2 } = (10⋅9)/(2⋅1)

= 4(45x^{0})

Hence the constant term is 180

**Example 4 :**

Find the constant term in the expansion (2x^{2} + 1/x)^{12}

**Solution :**

**T _{(r+1)} = ^{n}c_{r} x^{(n-r)} a^{r}**

Comparing the given expression with the form (x + a)^{n}, we get x = 2x^{2}, a = 1/x and n = 12

T_{(r+1)} = ^{12}c_{r} (2x^{2})^{(12-r)} (1/x)^{r}

= ^{12}c_{r} (2^{12-r})(x^{2})^{(12-r)} (x^{-r})

= 2^{12-r [}^{12}c_{r} x^{(24-3r)}^{]}

Let T_{r + 1} be the constant term

24 - 3r = 0 ⇒ r = 8

T_{r + 1} = T_{8 + 1}

= 2^{12-8 [}^{12}c_{8} x^{24-3(8)}^{]}

= 2^{4 }(495) x^{0 } ==> 7920

Hence the constant term is 7920.

After having gone through the stuff given above, we hope that the students would have understood, "Use the binomial theorem to find the indicated coefficient or term".

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